Please Hurry !!

Mathematics

1 answer:

Nuetrik [128]3 years ago

4 0

Answer:

d = 711 miles

Step-by-step explanation:

The speed s of the car is given by :

$s=3\sqrt{d}$ ...(1)

Where

d is the stopping distance

We need to find the stopping distance of a car travelling at 80 mph.

Rearranging equation (1) for d.

$s^2=9d\\\\d=\dfrac{s^2}{9}\\\\d=\dfrac{80^2}{9}\\\\d=711.1\ miles$

d = 711 miles

So, the required stopping distance is 711 miles.

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why write a quadratic function whose graph has the given characteristics vertex :(2,3) point on graph (0,-1)

GaryK [48]

<h3>Answer:</h3>

f(x) = -(x -2)² +3

<h3>Step-by-step explanation:</h3>

We can fill in the vertex (h, k) values immediately in the vertex form ...

... f(x) = a(x -h)² +k

To find the value of a, we solve the equation for a at some point other than the vertex. The given point is (0, -1), so we can use that:

... -1 = a(0 -2)² +3

... -4 = 4a . . . . . . . . . subtract 3, simplify

... -1 = a . . . . . . . . . . . divide by 4

Now, we know the function is ...

... f(x) = -(x -2)² +3

5 0

3 years ago

) An instructor gives his class a set of 18 problems with the information that the next quiz will consist of a random selection

RSB [31]

Answer:

The probability the he or she will answer correctly is 1.5%

Step-by-step explanation:

In all, there are 18 problems. In this question, the order of which the problems are sorted for the quiz makes no difference. For example, if the question A of the quiz is P1 and question B P2, and question A P2 and question B P1, it is the same thing.

There are 18 problems and 9 are going to be selected. So, there is going to be a combination of 9 elements from a set of 18 elements.

A combination of n elements from a set of m objects has the following formula:

$C_{(m,n)} = \frac{m!}{n!(m-n)!}$

In this question, m = 18, n = 9. So the total number of possibilities is:

$T_{p} = C_{(18,9)} = \frac{18!}{9!(18-9)!} = 48620$

Now we have to calculate the number of desired outcomes. This number is a combination of 9 elements from a set of 13 elements(13 is the number of problems that the student has figured out how to do).

Now, m = 13, n = 9. The number of desired possibilities is:

$D_{p} = C_{(13,9)} = \frac{13!}{9!(13-9)!} = 715$